Results 1 to 6 of 6

Thread: Chain Rule?

  1. #1
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1

    Chain Rule?

    Given that $\displaystyle x = e^t$, find $\displaystyle \frac{dy}{dx}$ in terms of $\displaystyle \frac{dy}{dt}$ and show that

    $\displaystyle \frac{d^2y}{dx^2} = e^{-2}\left(\frac{d^2y}{dt^2} -\frac{dy}{dt}\right)$.



    EDIT: Sorry, It is a typo, it should be $\displaystyle \frac{d^2y}{dx^2} = e^{-2t}\left(\frac{d^2y}{dt^2} -\frac{dy}{dt}\right)$.
    Last edited by Simplicity; Jun 15th 2008 at 02:49 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Feb 2008
    Posts
    40


    I remember doing this question!
    *dives into the past papers*




    I also remember doing it wrong and teacher correcting me

    I can do the first part though...

    Using the chain rule, we know:
    $\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{dy}{dt} . \frac{dt}{dx}$

    $\displaystyle x=e^t \\ \Longrightarrow \\ \frac{dx}{dt}=e^t \\ \Longrightarrow \\ \frac{dt}{dx}=\frac{1}{e^t}$

    and if you substitute that in, you'll get this:

    $\displaystyle \frac{dy}{dx}=\frac{1}{e^t}.\frac{dy}{dt}$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1
    Quote Originally Posted by Silver View Post


    I remember doing this question!
    *dives into the past papers*




    I also remember doing it wrong and teacher correcting me

    I can do the first part though...

    Using the chain rule, we know:
    $\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{dy}{dt} . \frac{dt}{dx}$

    $\displaystyle x=e^t \\ \Longrightarrow \\ \frac{dx}{dt}=e^t \\ \Longrightarrow \\ \frac{dt}{dx}=\frac{1}{e^t}$

    and if you substitute that in, you'll get this:

    $\displaystyle \frac{dy}{dx}=\frac{1}{e^t}.\frac{dy}{dt}$
    Yes, I got that too. But it's the next stage that is tricky. How do I differentiate that:

    $\displaystyle \frac{dy}{dx}=\frac{1}{e^t}.\frac{dy}{dt}$

    ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    Quote Originally Posted by Air View Post
    Yes, I got that too. But it's the next stage that is tricky. How do I differentiate that:

    $\displaystyle \frac{dy}{dx}=\frac{1}{e^t}.\frac{dy}{dt}$

    ?
    $\displaystyle \frac{d^2 y}{dx^2} = \frac{d}{dx} \left( e^{-t} \, \frac{dy}{dt} \right)$


    $\displaystyle = e^{-t} \frac{d}{dx} \left( \frac{dy}{dt} \right) + \frac{dy}{dt} \frac{d}{dx} \left( e^{-t} \right) $

    using the product rule


    $\displaystyle = e^{-t} \frac{d}{dt} \left( \frac{dy}{dt} \right) \, \frac{dt}{dx}+ \frac{dy}{dt} \, \frac{d}{dt} \left( e^{-t} \right) \frac{dt}{dx}$

    using the chain rule


    $\displaystyle = e^{-t} \, \frac{d^2y}{dt^2} \, e^{-t} + \frac{dy}{dt} \, \left(-e^{-t} \right) \, e^{-t}$


    $\displaystyle = e^{-2t} \, \frac{d^2y}{dt^2} - \frac{dy}{dt} \, e^{-2t} .....$
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hi,

    I'm not really sure because I can't get to your equation...
    And sorry for the "d", it takes too long time to straighten them...

    $\displaystyle {\color{red}\frac{dy}{dx}}=e^{-t} \frac{dy}{dt}$


    You're looking for $\displaystyle \frac{d^2y}{dx^2}$

    -----------------------

    $\displaystyle \begin{aligned} \frac{d^2y}{dx^2} &=\frac{d}{dx} \left({\color{red}\frac{dy}{dx}}\right) \\ \\
    &=\frac{d}{\color{green}dx} \left(e^{-t} \frac{dy}{dt}\right) \end{aligned}$

    Now, $\displaystyle x=e^t \implies {\color{green}dx}=e^t dt$ (we have to do it because the variable is t in the brackets, not x).

    Replacing :

    $\displaystyle \begin{aligned} \frac{d^2y}{dx^2} &=\frac{d}{\color{green}dx} \left(e^{-t} \frac{dy}{dt}\right) \\
    &=\frac{d}{e^t dt} \left(e^{-t} \frac{dy}{dt}\right) \\
    &=e^{-t} \frac{d}{dt} \left(e^{-t} \frac{dy}{dt} \right)
    \end{aligned}$

    Using the product rule :

    $\displaystyle \begin{aligned} \frac{d^2y}{dx^2} &=e^{-t} \left(-e^{-t} \frac{dy}{dt}+e^{-t} \frac{d^2y}{dt^2} \right) \\
    &=e^{-2t} \left(-\frac{dy}{dt}+\frac{d^2y}{dt^2}\right) \end{aligned}$




    Edit : oh I see, you made a typo, Air

    Re-edit : too fast for me Mr F
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jun 2008
    Posts
    20
    I think You have written the second part wrong. It should be:
    $\displaystyle
    \frac{d^2y}{dx^2} = e^{-2t}\left(\frac{d^2y}{dt^2} -\frac{dy}{dt}\right)
    $
    Solution:
    $\displaystyle
    \frac{dy}{dx}=\frac{1}{x}\frac{dy}{dt}
    $
    $\displaystyle
    \frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\ right)=\frac{d}{dx}\left(\frac{1}{x}\frac{dy}{dt}\ right)=\frac{1}{x}\frac{d}{dx}\frac{dy}{dt}-\frac{1}{x^2}\frac{dy}{dt}=\frac{1}{x}\frac{d}{dt} \frac{dy}{dx}-\frac{1}{x^2}\frac{dy}{dt}=\frac{1}{x^2}\frac{d^2y }{dt^2}-\frac{1}{x^2}\frac{dy}{dt}
    $
    $\displaystyle
    =e^{-2t}\left(\frac{d^2y}{dt^2} -\frac{dy}{dt}\right)
    $

    Regards,
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 9
    Last Post: Nov 9th 2010, 01:40 AM
  2. Chain Rule Inside of Chain Rule
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Oct 22nd 2009, 08:50 PM
  3. Replies: 5
    Last Post: Oct 19th 2009, 01:04 PM
  4. Replies: 3
    Last Post: May 25th 2009, 06:15 AM
  5. Replies: 2
    Last Post: Dec 13th 2007, 05:14 AM

Search Tags


/mathhelpforum @mathhelpforum